Let \(\pi_{1}\) and \(\pi_{2}\) be two \(k\)-dimensional planes in symplectic \(\mathbb{R}^{2 n}\). Is it always possible to carry \(\pi_{1}\) to \(\pi_{2}\) by a symplectic transformation? How many classes of planes are there which cannot be carried one into another?

A symplectic transformation is a linear transformation that preserves the symplectic form of a given space. In other words, if we have a symplectic space (V, ω), a symplectic transformation T : V → V must satisfy the condition: \(T^{*}\omega = \omega\) where \(T^{*}\) is the transpose of the matrix representing T in a given basis, and ω is the symplectic form.

Since we are given a symplectic \(\mathbb{R}^{2n}\) space, we know that there is a symplectic basis, \({a_1, b_1,..., a_n, b_n}\), which generates the whole space. For each k-dimensional plane, \(\pi_i\), where \(1 \leq i \leq k\), we can find a set of linearly independent vectors, \({v_1, v_2,..., v_k}\), such that: \(\pi_i = \operatorname{Span}(v_1, v_2,..., v_k) \subset \mathbb{R}^{2n}\)

Since a symplectic transformation preserves the symplectic form, we need to ensure that the symplectic form of vectors in \(\pi_{1}\) can be carried to the symplectic form of vectors in \(\pi_{2}\). Specifically, we must check the conditions: \(\omega(v_i, v_j) = \omega(T(v_i), T(v_j))\) for every \(1 \leq i,j \leq k\).

It is possible to carry \(\pi_{1}\) to \(\pi_{2}\) by a symplectic transformation if and only if the symplectic form of vectors in \(\pi_{1}\) can be transformed into the symplectic form of vectors in \(\pi_{2}\). Under these conditions: 1. The dimensions of \(\pi_{1}\) and \(\pi_{2}\) must be equal, as symplectic transformations are linear transformations. 2. For every \(1 \leq i,j \leq k\), \(\omega(v_{1_i}, v_{1_j}) = \omega(T(v_{1_i}), T(v_{1_j}))\) for the vectors \(v_{1_i},v_{1_j} \in \pi_{1}\) and \(T(v_{1_i}),T(v_{1_j}) \in \pi_{2}\). If these conditions are met, then a symplectic transformation exists that carries \(\pi_{1}\) to \(\pi_{2}\).

To determine the number of classes of planes that cannot be carried into one another, we need to identify how many different symplectic forms are there for k-dimensional planes in \(\mathbb{R}^{2n}\). A k-dimensional plane can be uniquely identified by the value of its symplectic form. The symplectic form takes values in the intervals \([0, k]\). Therefore, there is a total of \((k+1)\) different symplectic forms for a k-dimensional plane in \(\mathbb{R}^{2n}\). Now, we need to find out how many of these forms cannot be carried into one another by a symplectic transformation. Recall that symplectic transformations must satisfy the conditions specified in Step 4. It is clear that if two k-dimensional planes have different symplectic forms, they cannot be carried into one another. So, the number of classes of k-dimensional planes that cannot be carried into one another is equal to the number of distinct symplectic forms, which is \((k+1)\). In conclusion, it is not always possible to carry a k-dimensional plane \(\pi_{1}\) to another k-dimensional plane \(\pi_{2}\) by a symplectic transformation. There are \((k+1)\) classes of k-dimensional planes that cannot be carried into one another.